Quick question on prime focus imaging

[Keithp]

Hi all,

The other night I finally managed to achieve focus using my canon 500d on a skywatcher 250px (f4.7).

Just so I've got some idea for comparison purposes with my fuji s5700.

What sort of magnification am I getting with that setup and how do you work it out? As I progress with this, first I intend imaging as many targets as I can with this setup, but then move on to using 2" tele-extends to try to go deeper.

Regards

All

Keithp

[Merlin66]

At prime focus the "magnification" doesn't mean anything. The image size and the FOV is related to the focal length and the size of the CCD chip. The resolution is defined by the pixel size.

So, at prime focus you'll be operating at f4.5, with a fl of 1125mm.

With a 500D the chip size is 22.2mm x 14.8mm, pixel size 6.4 micron.

This give a FOV 46 x 69 arc mins, and a scale of 1.2 arc sec/ pixel, so the resolution would be about 2.5 arc sec ( you need a minimum of two pixels in the image)

hope this helps.

[Keithp]

Thanks for the reply Merlin, I guess that is the bit I can't seem to get my head round. The focal length of the scope (and please correct me if I'm wrong in my understanding of this) is the length of the cone of light bouncing back off the primary mirror. If I had a flat primary mirror then the image I would be seeing would be the portion of the view that is reflected by the secondary, totally unmagnified.

But surely the cone that is being reflected back up and off the secondary is showing me a closer (bigger) portion of view of the field of view and is magnified?

When we're talking about magnification with lenses and focal lengths ie the scope is 1200 a lense might be 25 and 1200/25 gives a magnification of 48. So that's the value brought back by the focal length of the mirror times the fl of the eypeice. If we take away the eyepeice then are we not left with a magnified view purely from the primary mirror?

If we look at that view with different sized chips are you saying the different size of the chips will make that view look bigger or smaller depending on the size of the pixels. Is not the picture a constant?

Regards

Keithp

[Merlin66]

The secondary mirror in a newtonian is flat and doesn't do anything other than bend the cone of light to the side of the tube.

Yes, there is some "magnification" from the primary mirror, but it's a bit like a shaving mirror.. our normal viewing distance, when reading a newspaper etc is about 250mm, so if you used your primary mirror for shaving it would probably "magnify" 1200/250 = 4.5 times

When it comes to imaging, the focal length of the system is what it's all about ( yes, the focal ratio comes into it - but not at the moment!)

The size of an object IN THE FIELD OF VIEW is dependent on the effective focal length. Jupiter would appear very small with a fl of 1200mm but three times bigger when you use a fl of 3600.

The FOV is limited by the size of the CCD chip, if you used a webcam you'd only record about a 1/5 of the FOV what you'd get with the canon.. So Jupiter would appear the same size ( say at 3600mm fl) but with less sky around it in a webcam than in the Canon. The pixels ONLY affect the resolution and detail you can record in the image.

Hpe this helps.

[Keithp]

Ah that's what I was after though:hello2:. The magnification of the image purely by the mirrors in the tube is the focal length divided by the appeture.(I assume this is only relevent to reflectors) This was just to give some idea so I appreciate what I would have with various length barlows/telextenders and so on. I am now finding this imaging malarky as engaging as pure viewing.

Thanks Merlin

Regards

Keithp

[Merlin66]

Keith,

That's the wrong conclusion!!!

The primary mirror makes an image, the size of which is dependent on the focal length....

The eyepiece then acts as a magnifying lens to "magnify" this primary image.

The more powerful the eyepiece ( lower fl length ie 6mm v's 25mm) the greater the magnification. The layout of the optics give the formula :

Magnification = Focal length of the primary/ focal length of the eyepiece.

Sorry if I confused you.

[Keithp]

"The primary mirror makes an image, the size of which is dependent on the focal length...."

So that brings me back to my first question, when I attach the camera at primary focus what is the magnification of the image I am seeing, as it is only the image being projected by the mirrors. And how is this worked out?

Sorry Merlin if this is a miss grasped concept, but I'm trying to understand what effects I would put on the image using various barlows and telextenders etc.

Regards

Keithp

[Merlin66]

keith,

It's just terminology. there's NO magnification at the primary focus. Magnification is a poor word to use. The longer the focal length, the larger the primary image.

So, if you subsequently use say, a x 2 barlow; all it does is to increase the focal length ( from 1200 to 2400mm) and double the primary image size

[Keithp]

I should re-title this as 'Non-Quick question on prime focus imaging':o

Let me think about that for a while, thanks Merlin.

Regards

Keithp

[Milamber]

Standard wisdom is that dropping a ccd in the imaging train is eq to a 6 to 8mm EP Keith... from that you can figure your "mag" but it's really a misnomer, just because you are seeing something does not mean it's magnified. You are looking (in prime imaging) at the image focused on the ccd as seen by the primary. With no eyepiece in the way there is no magnification *of the image*. Think of it like this. With a 1000 focal length, you need a 10mm eyepiece to get 100x yes? that gives you 100x "mag", right? with a 500 focal length you would need a 5mm eyepiece to get the same 100x "mag", true? (that 10mm would only give you 50x). With me so far? How can two different eyepieces give the same magnification? Or, how can the same eyepiece give two different magnifications? The answer is as Merlin was trying to say - it's a misnomer to call it magnification since that implies a standard, and with so many different telescopes there cannot *be* a standard. Hence, you see the image as created by the mirror/objective/whatever and it is not magnified, it's just the image produced by the optics currently in use.

Arthur

[arad85]

Another way to think of it..

What you are doing with different focal length 'scopes is generating (at the focal point) an image which is a certain size. With an eyepiece, this then gets magnified by the eyepiece before it goes into your eye, depending on the strength (i.e. focal length - think of them as magnifying glasses) of the eyepiece.

With prime focus photography, there is no eyepiece to do the magnification. What you are doing is putting the imaging element at the point where the sharp picture is generated. Now, the imaging element is fixed in size, so the area of the sky that gets projected onto the imaging element will depend solely on the focal length of the scope. The longer the tube you are looking down, the smaller area of the sky will produce the same image size (the size of the imaging element). So effectively you get higher "magnification" with a longer tube length. However....

The "magnification" you get will then depend on two things - the resolving power of the scope and the physical dimensions of the pixels. Assuming the same sized imaging elements (most dSLRs are APS-C I believe) then more pixels will mean (assuming the 'scope can resolve to beyond the limit of the pixels) that more pixels will be spread over the same amount of sky, so you get a higher resolution not magnification (each pixel is able to resolve a smaller area of the resulting projected image).

I think there are at least two questions here:

- what's the relationship between focal length and the area of the sky that can be seen on an APS-C dSLR?

- How do I work out what my resolution (in arc seconds) is per pixel given the number of pixels on my sensor?

[Keithp]

By jove I think I got it:hello2:

Not decrying what Merlin and Milamber have added I think Arad85 has cleared it up for my muddled brain.

What you're all saying (and correct me again if wrong) is the image is a standard for the focal length of the scope, all f4.7 scopes will generate the same size image, all f10 scopes will generate the same size image, you then use different fl sized eps to magnify that image, but at the focal length of the scope...yes? The magnification is the application of the eyepeice to the focal length image, and that can vary scope to scope. I understand that....

But... the image of an f4.7 scope compared to an image of an f10 scope without eyepeice is...larger? yes/no?. So what is that image size difference compared to a flat mirror...in the case of the f4.7 scope is it 1200/250 (my scope) times bigger than a flat reflected image?

Regards

Keithp

[Merlin66]

You were doing well when you said "What you're all saying (and correct me again if wrong) is the image is a standard for the focal length of the scope" then next part is where you go wrong.

It's the focal length not the focal ratio that determines the primary image size; a 200mm f5 (1000mm fl) will give a smaller image than a 400mm f5 (2000mm fl)

You can't compare the image with that of a "flat mirror" it's apples and pears!!

[Keithp]

Bear with me...it will sink in as soon as I understand it properly.

The focal length is the length of the cone of light? yes/no?

That cone of light comes from different size appetures, so the first one (below) gives a smaller area of sky to the second one, but any objects in the field of view are the same size as they are the same focal length? yes/no?

Assume the one on the left is 200mm appeture the one on the right is 400mm and both are fl 1200 (is that possible?)

[------] [-------------]

(no good I tried to show two cones below each of the above but formatting wouldn't let me, imagine two cones extending below both lines, the same length, but one wider that the other)

If I put a 20mm (fl) ep what magnification do I get on the one on the left compared to the one on the right if they are the same focal length? The same? but on a larger image from the one on the right?

Regards

Keithp

[Merlin66]

"The focal length is the length of the cone of light? yes/no?" -YES

Your example with the different apertures....

Its ONLY the focal length, NOT the aperture which influences the FOV.

Both examples CAN have the same focal length, but DIFFERENT focal ratios:

The first has a focal ratio = 1200/200 = f6

the second, 1200/400 = f3

A 20mm eyepiece will give the SAME magnification on each telescope = 1200/20 = x600

[Keithp]

Merlin, my gratitude for bearing with me on this, I'm still not grasping the concept.. I have a preconceived idea of what I'm seeing that is not in line with what you're saying.

'That cone of light comes from different size appetures, so the first one (below) gives a smaller area of sky to the second one, but any objects in the field of view are the same size as they are the same focal length? yes/no?'

Is that correct? yes/no?

'Its ONLY the focal length, NOT the aperture which influences the FOV.'

Surely the larger aperture returns a larger field of view, but brought to the same size as a smaller aperture with the same focal length?

ie if I was looking at Jupiter, no lenses just the mirrors, then I would see more space around Juptiter through the larger apeture scope than the smaller apeture scope?

Regards

Keithp

Ps I really want to nail this one so I can understand what I'm seeing at prime. I can understand the camera with (new can of worms) depth of field ...no. no.. no not going there until I get this sorted:o

[Merlin66]

'That cone of light comes from different size appetures, so the first one (below) gives a smaller area of sky to the second one _NO

That's were you're going wrong; you're thinking of the cone of light reflected from the primary as having some effect on the FOV it doesn't.

[Keithp]

Merlin, many thanks...It's too late at night for me to digest this without posting rubbish...maybe some of what I've already posted is that:eek:

I'm going to think on what you said:headbang:

I have some fundamentals to sit with and sort out.. if you know what I mean.

Still, doesn't take away from getting a focus at prime with the new kit tho:)

Heh heh, got me first +4 secs exposure on my other thread..

Again, many thanks Merlin... we engage tomorrow:)

Regards

Keithp

Ps I notice we got a few viewing on this....anybody else fancing droping in with opinions..

[arad85]

I think the bit you are missing here is that a longer focal length system actually bends light LESS. Think of it as a very large camera lens (which is actually what it is!!). If you have a 400mm diameter lens capable of focusing at 2000mm you have an F5 system. You could also stop this down (as a camera does) to 200mm with a 200mm tube underneath the lens. It's still a 2000mm FL system as it's using the same lens, just stopped down to F10 with the tube. The magnification is still the same, just you need much longer exposure (4x) to get the same brightness (as you have less light falling on your sensor). What the extra diameter of the lens gives you here is more light gathering.

A telescope of a certain focal length generates an image (inside the scope - this is known as the focal plane) a certain size, depending ONLY on the focal length of the scope. In fact, the formula for working out where a point will appear on the focal plane is given by the formula fl * tan (angle) where fl is the focal length of the scope and angle is the amount the point you are looking at is off the central axis of the lens. You should be able to see that for stars, the angle will always be constant for any two stars, so the only variable is the focal length of the scope. Now, you are putting your image sensor at the focal plane, so if you have a 'scope with 2x the focal length, the two stars will be 2x the distance apart on the sensor. Note this is totally irrespective of the aperture of the scope - it ONLY depends on the focal length.

I mentioned above, that aperture just gave you more light - that is true, but it also gives you more angular resolution. For a coherent explanation (i.e. I'm still getting my head around it), have a look at: Angular resolution - Wikipedia, the free encyclopedia You'll note that the aperture is on the bottom of the equation which means the larger the aperture, the more angular resolution you have.

So, to sum up. The size of the image from the 'scope on the sensor is determined by one thing - the focal length. 2x the focal length gives 2x the image size (this is logical given magnification of a telescope is determined by fl scope/fl eyepiece - 2x fl scope = 2x magnification). Aperture gives you two things - faster imaging given the same focal length AND increase in angular resolution.

An example. Let's take two 'scopes with a focal length of 1000mm, and one with an aperture of 100mm (A) and the other an aperture of 200 mm (. We can say the following things if you image:

• The image from both scopes will be the same size (i.e. stars are separated by the same distance on both images)
  
• Scope A will take 4x the exposure to get the same brightness as scope B this is because scope B is twice as "fast" as scope A)
  
• The image from scope B will be twice as "sharp" as that from scope A (i.e. you can resolve more detail if the sensor is capable of it). By this I mean that scope A will focus a star to an area 4x that of the same star focused by scope B (assuming perfect optics).
  
• There is no "magnification" factor in either scope as you are not passing the generated image through a magnifying glass (eyepiece) to magnify the resulting image. What you do get are stars that are separated further apart the longer the focal length gets and then the "magnification" then depends on the medium used to display that image (you could put it on a 6x4 print, 10x7 print, on screen - the stars will look bigger/smaller depending on how they are rendered).
  

Does that help?

[arad85]

You may also find this link useful:

Choosing a camera for astrophotography

[Keithp]

Arad85, well done, you're last but one explanation tied it together I can understand what you're on about now, and the other explantions from others fitting the numbers in....Phew took a while, but I got it:(

Thanks to everyone who contributed to this thread:hello2:

Regards

Keithp

[arad85]

You're welcome Made me understand what's going on here too !!

[themos]

Start with the easy things first: no lens.

Get a pinhole camera, what magnification does it produce?

Well, a pinhole camera produces a clear image no matter how far the sensor is from the pinhole.

If you shoot a flower that is 1 meter away and you put the sensor at 1m from the pinhole you are going to get a 1:1 picture of the flower on your sensor.

If you shoot the moon that is 380,000 km away and you put the sensor at 1 meter again, you will get a picture of the moon that is 380,000,000 times SMALLER than the moon.

If you put the sensor at 2 meters from the pinhole, you'll get a picture that is only 190,000,000 times smaller than the moon.

So much for "magnification"!

Magnification, then, depends on

a ) how far you put the sensor from the pinhole.

and

how close you choose to hold the picture to your eyes.

This last fact makes "magnification" a useless concept for images.

The first fact is what people have been trying to explain to you: larger pinhole-sensor distance makes bigger image.

Lenses work just like pinholes, except that you have to put the sensor at a fixed distance from the hole (that's the focal length of the lens) and the glass conspires to send all the light from the equivalent myriad of pinholes to THE SAME PLACE on the sensor. It can ONLY do that when the sensor is at the focal length (we are thinking of light that originates at effective infinity, here, which in practice means many times further away than the focal length).

[Keithp]

Oh oh, Themos....you got me thinking again....

....let me think about that..might be the same concept...sounds deceptively the same..but subtly different.....hmmmm

back later.......after some thought.

, Still got the remote shutter control coming for the weekend....:hello2:longer exposers......and trails....he he maybe.

Local forecast....clear sky this Friday night/Saturday:cool:

Regards

All

Keithp

[themos]

Think about it, if I told you that the magnification of your imaging setup was 60x would that help you decide if the full moon could fit in your frame?

No, it wouldn't.

What WOULD help is if you could calculate how many degrees of sky your camera frame covers. The moon is about 1/2 degree.

Back to our pinhole camera. We can tell very quickly how many degrees of sky a sensor of size D (in mm) would see if placed a distance F (in mm) from the pinhole. Isosceles triangle with base D and height F, what's the subtended angle z? Well, tan(z/2) = (D/2)/F. Plug some numbers in (D=20mm, F=1000mm)

tan(z/2) = 20/2 / 1000 = 10/1000 = 0.01

My windows XP calculator gives z/2 = 0.573 degrees to 3 digits, which means the whole angle is twice that: 1.146 degrees. For such small angles, you might as well take (z/2)=(D/2)/F, provided you remember that z will be given in radians and you need to multiply by 180 and divide by pi to get degrees. 0.02*180/pi also gives 1.146 to 3 digits.